Question

The general solution of the differential equation $(1+\tan y)(dx-dy)+2xdy=0$ is

• A. $x(\sin y+\cos y)=\sin y+c{e}^y$
• B. $x(\sin y+\cos y)=\sin y+c{e}^{-y}$
• C. $y(\sin x+\cos x)=\sin x+c{e}^x$
• D. None of these

Answer 1

Answer: (B)

We have, $(1+\tan y)(dx-dy)+2xdy=0$
$\Rightarrow (1+\tan y)dx=(1+\tan y-2x)dy$
$\Rightarrow \frac{dx}{dy}+\frac{2}{1+\tan y}x=1,$ which is linear differential equation.
I.F. $=e^{2\int \frac{dy}{1+\tan y}}=e^{\int \frac{2\cos y}{\sin y+\cos y}dy}$
$=e^{\int \left(1+\frac{\cos y-\sin y}{\sin y+\cos y}\right)dy}=e^{y+\log (\cos y+\sin y)}$
$=(\cos y+\sin y)e^y$
So, the solution is
$x{e}^y(\sin y+\cos y)=\int e^y(\sin y+\cos y)dy+c$
i.e. $x{e}^y(\sin y+\cos y)=e^y\sin y+c.$
i.e. $x(\sin y+\cos y)=\sin y+c{e}^{-y}$